iitjee two yrs comprehensive crs for iit jee 2013 (csp,cjp,cjp+) mathematics target1 chapter2...
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8/4/2019 Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2 Sectiona
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Chapter - 2 (Section A:Objetive Type Single Choice)
Q.No. Solutions
1. Answer (2)
cos1°.cos2°……cos179° = 0
(as there will be a term cos 90° in this series whose value is zero)
2. Answer (1)
13sec ,0
5 2
2
2 3cot
4 9 sec 1
=
52 3
1212
4 95
( 2 2sec 1 tan )
1213
5
=3/4 15
88/5 352
3. Answer (1)2(cosec .cosec cot .cot ) A B A B 2(cosec .cot cosec .cot ) A B B A
2 2 2 2 2 2cosec (cosec cot ) cot (cosec cot ) A B B A B B
2 2cosec cot 1 A A
4. Answer (2)
5 99sin , sin
13 101 A B
cos( ) cos .cos sin .sin A B A B A B
513
12
99
20
101
=12 20 5 99
. .13 101 13 101
=255
1313
A + B + C = A + B = – C
or cos ( A + B) = cos ( – C )
cos C = – cos ( A + B)
=255
1313
8/4/2019 Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2 Sectiona
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Q.No. Solutions
5. Answer (2)
tan cot a , 4 4tan cot ? 4 4tan cot = 4 2 2 2 2(tan cot ) 4 tan .cot .(tan cot ) 6 tan .cot
= 4 24{(tan cot ) 2 tan .cot } 6a
= 4 24 2a a
6. Answer (1)
cos sin 3a b sin cos 4a b
Squaring and adding, a2
+ b2
= 25
7. Answer (1)
cos , sina b c d
cos ,sinb d
a c
Squaring and adding,2 2
2 21
b d
a c
2 2 2 2 2 2b c a d a c
8. Answer (3)
2
sin10
x x , [ , ] x
Clearly the total number of points of intersection is 3.
9. Answer (3)2 2
max min(3 sec ) (4 tan ) 2 4 2 x y
10. Answer (1)
tan ,
p
q
sin cos
?sin cos
p q
p q
Divide numerator and denominator by cos
2 2
2 2
tan
tan
p q p q
p q p q
11. Answer (3)sin1° < sin 1 is correct
As a 1 radian =180
degree 13.14
sin is increasing in 1st
quadrant
8/4/2019 Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2 Sectiona
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Q.No. Solutions
sin1° < sin 1
12. Answer (2)
sin1 cos2 tan3 cot 4 sec 5 cosec 6
= negative
13. Answer (1, 4)Both the options are correct
1
2
tan 1 > tan 2 (from graph)
14. Answer (1)
3sin 4cos 5 4sin 3cos Squaring and adding,
29 16 25 0
15. Answer (2)
sec tana c d sec tana d c sec tanb d c sec tanb c d Squaring and adding,
2 2 2 2 2 2 2( )sec (1 tan ) (1 tan )a b d c
2 2 2 2a b c d
16. Answer (3)
1 sin sec tan( )
1 sin sec tan
f
2(sec tan ) |sec tan |
Now f () = g ()
If sec – tan 0
1 sin0
cos
cos > 0
,2 2
17. Answer (1)
12 2
2r l r where r is radius of the circle and l is the length of the arc i.e. l r
2r r r
8/4/2019 Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2 Sectiona
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Q.No. Solutions
( 2) radian
18. Answer (1)
1sin , 2
4 42
1cos ,
2 3 3
tan 3 , 23 3
Sum of all possible values of . , 2 22
8 –3 3
19. Answer (2)Let the angles be , ,d d
3 180 60 180 60
d
d
d = –30°Hence the angles are 30°, 60°, 90°.
20. Answer (1)2 3sin sin sin 1 0 x x x
2(sin 1)(sin 1) 0 x x
sin x = –1
cos x = 06 4 2cos 4cos cos 0 x x B x
21. Answer (1)
12tan 5 0 A , 5cos 3 0B
5
tan12
A ,3
cos5
B
ABCD is a cyclic quadrilateral
A + C = , B + D = C = – A, D = – B
cos –cosC A tan tan D B
=12
13 =
4
3
Equation whose roots are cos C and tan D is 2 12 4 12 4 013 3 13 3
x x
239 16 48 0 x x
22. Answer (2)
Let 4 4sin cos and 8 8sin cos
max a , max = b, a – b = ?
2 2 2 2 2(sin cos ) 2sin cos
211 sin 2
2
8/4/2019 Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2 Sectiona
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Q.No. Solutions
max = a = 1Similarly, max = b = 1
a – b = 0
23. Answer (2)
cos sec 2 , [0, 2 ]
cos sec 1
8 8sin cos = 0 + 1 = 1
24. Answer (1)
f () = 2 24cos 24sin .cos 6sin
=
1 cos2 1 cos2
4 12sin2 62 2
= 5cos2 12sin2 1 f max = 13 – 1 = 12f min = – 13 – 1 = –14
Range of f () is [–14, 12]
25. Answer (3)
4(sin4 x – cos
4 x ) + cos
2 x = k
1 cos2
4( cos2 )2
x x k
4(sin2 x – cos
2 x ) + cos
2 x = k
4 – 7cos2 x = k
4
0 17
k
Then
26. Answer (3)
3sin
3
x x
x , 0 x
1 LHS 1 RHS 2 or RHS 2
No solution
27. Answer (4)
Let a be the side of dodecagon i.e. 3 1a
And let b the side of hexagon and r be the radius of the circle in AOC, AOD = 45°
sin15 AD
AO
3 1 3 1
2 2 2r
2r In ,
sin302
b
r
8/4/2019 Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2 Sectiona
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Q.No. Solutions
12 2 2
2b
28. Answer (1)
cos(cos )y x
max1 , when
2y a x
mincos1 when 0y b x
b = cos a
29. Answer (2)
2sin
1 sin cos
Ak
A A
,1 sin cos
?
1 sin
A A
A
2 2
2sin .(1 sin cos )
(1 sin ) cos
A A Ak
A A
2 2
2sin .(1 sin cos )
(1 sin ) (1 sin )
A A Ak
A A
=2sin .(1 sin cot )
(1 sin ).{(1 sin ) (1 sin )}
A A A
A A A
=1 sin cos
1 sin
A A
A
30. Answer (1)2sin sin 1 x x
2sin cos x x 12 10 8 6cos 3cos 3cos cos 1 x x x x
= 6 6 4 2cos .(cos 3cos 3cos 1) 1 x x x x
= 2 2 3{cos (1 cos )} 1 x x
= 3{sin (1 sin )} 1 x x
= 2 3 3{sin sin } 1 (1) 1 0 x x
31. Answer (3)
, satisfy 2cos cos 0 x a x b and 2sin sin 0 x p x q cos cos a … (1)
cos .cos b … (2)
sin sin p … (3)
sin .sin q … (4)
Squaring and adding (1) & (2)2 22 2(cos cos sin sin ) a p
2 22 2( )b q a p 2 2 2 2( )a p b q
32. Answer (1)
sec A and cosec A are the roots of 2 0 x ax b
8/4/2019 Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2 Sectiona
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Q.No. Solutions
sec cosec A A a , sec .cosec A A bsin cos
sin .cos
A Aa
A A
,
1sin .cos A A
b
sin cosa
A Ab
Squaring,
2
21 2sin .cos
a A A
b
2
2
21
a
b b
2 22b b a
or 2
( 2)a b b
33. Answer (1)29
115
cos2
r
r S
29
215
sin2
r
r S
S1 =29
15
cos2
r
r
=15 16 29
cos cos .... cos2 2 2
= (sin cos sin cos ) ..... (Sum of consecutive terms is zero and these are 15 terms in all)
S1 = (sin cos sin ) = cos
S2 =29
15
sin2
r
r
= – cos sin cos = sin
1
2
S
S=
cos
sin
= cot
34. Answer (3)
f () = 23 2sin cos
= 23 2sin 1 sin
= 2(sin 2sin 1) 5
= 25 (sin 1)
f max = 5f min = 1
max
min
f
f =
55
1
8/4/2019 Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2 Sectiona
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Q.No. Solutions
35. Answer (1) 2 2cos cos(33 ) sin sin(45 )
= 2 2cos ( cos ) sin ( sin )
= 2 2cos ( cos ) sin (sin )
cos cos(cos ) 1 2 2cos 1 cos (cos ) 1
and 2 20 sin (sin ) sin 1
2 2 2 2cos 1 cos (cos ) sin (sin ) 1 sin 1
Maximum value = 21 sin 1 At2
36. Answer (3)4 4sin cos 1
2 3 5
x x
5(3tan4 x + 2) = 6 sec
4 x
15tan4 x + 10 = 6 + 6tan
4 x + 12 tan
2 x
9tan4 x – 12tan
2 x + 4 = 0
(3tan2 x – 2)
2= 0
25
3
2 2tan3
x
2 22 3sin & cos
5 5 x x
8 8sin sin
8 27
x x
=16 81
8 625 27 625
=5 1
625 125
37. Answer (1)
2
2sec tan , (2 1)
2sec tan x x k x x x x
=2
2
1 tan tan
1 tan tan
x x
x x Let tan x = t
t 2(k – 1) + t (k + 1) + (k – 1) = 0
D = (k + 1)2
– 4(k – 1) 0
If k = 1, we get2
2
11
1
t t
t t
3k
2– 10k + 3 0
8/4/2019 Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2 Sectiona
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Q.No. Solutions
max625
2t where sin2( –) = 1
43. Answer (4)
44. Answer (3)
1 2 1 3 1 4
1 1 1
A A A A A A
Let the polygon be inscribed in a circle of radius ‘r ’ A
4A
3
A2
D
A1
Or
Each side subtends angle2
n
at the centre
1 2 2 3 3 4
2 AOA A OA A OA
n
In OAD,
1 2 sin2
A A AD r
n
1 2 2 sin A A r n
Similarly 1 3 1 4
2 32 sin & 2 sin A A r A A r
n n
1 1 1
2 32 sin 2 sin 2 sinr r r
n n n
3 2sin sin
1
2 3sin sin sin
n n
n n n
3 2sin sin
1
3sin 2sin cos sin
n n
n n n n
4 2 3 2
sin sin sin sinn n n n
4 3
sin sinn n
4 3( 1)k k
n n
4 30,k
n n
4 31,k
n n
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Q.No. Solutions
77n
n
45. Answer (2)
asin2+ bcos
2= m …(1)
bsin2+ acos
2= n …(2)
atan = btan(ii) …(3)
Divide (1) by cos2 we, get
a tan2 + b = m sec
2
2tanm b
a m
…(4)
Divide (2) by cos2, we get
Btan2 + a = nsec
2
2tann a
b n
…(5)
From (3), (4), (5)
2 2m b n aa b
a m b n
a2(mb – mn – b
2+ bn) = b
2(an – a
2– mn – am)
abm(a – b) + abn(a – b) = mn(a2
– b2)
abm + abn = mn(a + b)Divide both sides by abmn, we get
1 1 1 1
n m a b